### Author Topic: A question about zeroes of the Zeta function  (Read 2635 times)

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#### Xeda112358

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• Calc-u-lator, do doo doo do do do. ##### A question about zeroes of the Zeta function
« on: October 28, 2014, 01:24:11 pm »

The problem:
Is there an $s\in \mathbb{C}$ such that $\zeta(s)=\zeta(2s)=0$? If so, what is $\lim_{z\rightarrow s}{\frac{\zeta(s)^{2}}{\zeta(2s)}}$?

Okee dokee, I know this is getting dangerously close to the Riemann Hypothesis, but I swear that wasn't my original intent!

I was playing around with infinite sums and products and I wanted to figure out the sum form of $\prod_{p}{\frac{1+p^{-s}}{1-p^-s}}$. I started by knowing that $\prod_{p}{(1-p^{-s})}^{-1} = \zeta(s)$, and from previous work, $\prod_{p}{\sum_{k=0}^{n-1}{(p^{-s})^{k}}} = \frac{\zeta(s)}{\zeta(ns)}$). From that, I knew $\prod_{p}{1+p^{-s}} = \frac{\zeta(s)}{\zeta(2s)}$, thus I know the product converges when $\zeta(2s) \neq 0, \zeta(s) \in \mathbb{C}$. I knew convergence wasn't an issue (for the most part) so I expanded the product some (a reversal of Euler's conversion from $\zeta(s) = \sum_{k=1}^{\infty}{k^{-s}} = \prod_{p}{(1-p^{-s})}^{-1}$) and obtained:

$\prod_{p}{\frac{1+p^{-s}}{1-p^-s}}= \sum_{k=1}^{\infty}{2^{f(k)}k^{-s}}$,

where $f(k)$ is the number of prime factors of k. It turns out that this has been known since 1979 and after I had access to the internet, I figured out the typical symbol used for my $f(k)$ in this context is $\omega(k)$. So that was cool, and I could write that sum and product in terms of the zeta function as $\frac{\zeta(s)}{\zeta(2s)}\zeta(s) = \frac{\zeta(s)^{2}}{\zeta(2s)}$, so do with that what you will (I tried to use it to find the number of prime factors of a number n, but I didn't get anywhere useful). What I decided to pursue was when this function was zero. As long as $\zeta(s)=0, \zeta(2s)\neq 0$, we know that it is 0, but I haven't yet figured out if there is ever an instance where $s\in \mathbb{C}, \zeta(s)=\zeta(2s)=0$. If there is such an s, then the Riemann Hypothesis is false. However, if such an s does not exist, this says nothing about the Riemann Hypothesis .
Spoiler For Why the existence of an s would prove RH false:
It is simple: RH says that $\zeta(s)=0$ if and only if $Re(s)=\frac{1}{2}$. So if $\zeta(s)=\zeta(2s)=0$, assuming RH is true, then $Re(s)=Re(2s)=2Re(s)$, which would imply s=0, a contradiction to the Riemann Hypothesis which we just assumed was true!

As well, if there is such an s, I wanted to know if it could be one of those removable discontinuity thingies.

EDIT: fixed a tag and an incorrect restatement of the RIemann Hypothesis « Last Edit: October 28, 2014, 01:30:57 pm by Xeda112358 »